\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)

\(=\frac{34}{132}+\frac{1}{20}\)

\(=\frac{17}{66}+\frac{1}{20}\)

\(=\frac{203}{660}\)

\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\) 

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)

\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)

\(=\frac{3}{10}+\frac{1}{132}\)

\(=\frac{198}{660}+\frac{5}{660}\)

\(=\frac{203}{660}\)

#)Giải :

\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)

\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}=\frac{1}{3}-\frac{1}{21}=\frac{2}{7}\)

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}=7\left[\frac{1}{7}\left(\frac{7}{10.11}+\frac{7}{11.12}+...+\frac{7}{69.70}\right)\right]\)

\(C=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)=7\times\frac{3}{35}=\frac{21}{35}\)

#)Góp ý :

Ý D tương tự ý C

sửa đề: \(B=5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}\)

Ta có: \(A=\frac23+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}+\frac{194}{195}\)

\(=1-\frac13+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}+1-\frac{1}{143}+1-\frac{1}{195}\)

\(=7-\left(\frac13+\frac{1}{15}+\cdots+\frac{1}{195}\right)\)

\(=7-\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{13\cdot15}\right)\)

\(=7-\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{13}-\frac{1}{15}\right)=7-\frac12\left(1-\frac{1}{15}\right)\)

\(=7-\frac12\cdot\frac{14}{15}=7-\frac{7}{15}=\frac{98}{15}\) >6

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{7^2}<\frac{1}{6\cdot7}=\frac16-\frac17\)

Do đó; \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<1-\frac12+\frac12-\frac13+\cdots+\frac16-\frac17=1-\frac17<1\)

=>\(5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<5+1=6\)

=>B<6

mà A>6

nên B<A

1-1/2+1/2-1/3+1/3+1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130

cái này tính cái gì thế

ko hiểu

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)

\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)

\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)

\(\frac{6}{15}+\frac{3}{35}+\frac{6}{63}+\frac{6}{99}=\frac{2}{5}+\frac{3}{35}+\frac{2}{21}+\frac{2}{33}=\frac{247}{385}\)

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

 ~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~

mik cần tính nhanh nhé bn

Giúp mình với!

a,(2017-2).4+2=8062

b,(2017-8).10+8=20098
còn lại tự làm nha bạn